Codeforces Round 937 (Div. 4) B. Upscaling ..,##交替输出两次即可 1234567891011121314151617181920212223242526272829303132// By SnowDream#include<bits/stdc++.h>using namespace std;typedef long long ll;con

Codeforces Round 937 (Div. 4) C. Clock Conversion 代码： 12345678910111213141516171819202122232425262728293031323334// By SnowDream#include<bits/stdc++.h>using namespace std;typedef long long ll;co

Codeforces Round 946 (Div. 3) A.Phone Desktop 1234567891011121314151617181920212223242526272829303132333435// By SnowDream#include<bits/stdc++.h>using namespace std;typedef long long ll;const in

Codeforces Round 946 (Div. 3) B.Symmetric Encoding 12345678910111213141516171819202122232425262728293031323334353637383940// By SnowDream#include<bits/stdc++.h>using namespace std;typedef long l

Codeforces Round 937 (Div. 4) D. Product of Binary Decimals 本人使用了朴实无华的打表，从最大的开始代入遍历 代码： 123456789101112131415161718192021222324252627282930313233343536373839404142434445// By SnowDream#include<bits

Codeforces Round 944 (Div. 4) A. My First Sorting Problem 1234567891011121314151617181920212223// By SnowDream#include<bits/stdc++.h>using namespace std;typedef long long ll;const int N=1e6+10;i

Codeforces Round 946 (Div. 3) C.Beautiful Triple Pairs 代码的主要作用是计算在给定的多个测试用例中，每个测试用例的数组中特定模式（对和三元组）出现的次数，并对这些模式出现的次数进行加减操作以得到最终结果。 具体步骤如下： 读取测试用例数量 t。 对于每个测试用例，使用两个 map 分别记录三元组和二维对的出现次数。 读取数组长度 n 和数组元

Codeforces Round 934 (Div. 2) C. MEX Game 1 123456789101112131415161718192021222324252627282930313233343536373839404142434445#include<bits/stdc++.h>using namespace std;const int Max=1e6;int g[Ma

Codeforces Round 944 (Div. 4) B. Different String 123456789101112131415161718192021222324252627282930313233343536373839404142434445// By SnowDream#include<bits/stdc++.h>using namespace std;typed

Codeforces Round 944 (Div. 4) D. Binary Cut 遇到 ‘10’ 分割一次,‘01’ 分割 n-1(n为’01’出现次数) 次 123456789101112131415161718192021222324252627282930313233// By SnowDream#include<bits/stdc++.h>using namespace